I couldn't find this exact question, apologize if it's a duplicate.
I would like to show based only on the above definition of e, that this equality for $e^x$ holds, without going through showing it's a bijective function, has an inverse function ln, finding it's derivative and finding it's taylor series. I've tried to play with it, but except for x=-1 (with a proof that doesn't seem to be possible to generalize), I had no progress.
Is their a straightforward proof for that? Or does one has to go through the whole process mentioned above?
Extending the Limit Over the Integers to the Limit Over the Reals
For $\alpha\ge1$, $$ \underbrace{\left(1+\frac1{\lfloor\alpha\rfloor+1}\right)^{\large\lfloor\alpha\rfloor}}_{\color{#C00000}{\left(1+\frac{\large1}{\lfloor\alpha\rfloor+1}\right)^{-1}}\color{#00A000}{\left(1+\frac{\large1}{\lfloor\alpha\rfloor+1}\right)^{\large\lfloor\alpha\rfloor+1}}} \le\left(1+\frac1\alpha\vphantom{\frac1{\lfloor\alpha\rfloor}}\right)^{\large\alpha} \le\underbrace{\left(1+\frac1{\lfloor\alpha\rfloor}\right)^{\large\lfloor\alpha\rfloor+1}}_{\color{#C00000}{\left(1+\frac{\large1}{\lfloor\alpha\rfloor}\right)}\color{#00A000}{\left(1+\frac{\large1}{\lfloor\alpha\rfloor}\right)^{\lfloor\alpha\rfloor}}}\tag{1} $$ where the red terms tend to $1$ and the green terms tend to $e$ by the standard limit $$ \lim_{n\to\infty}\left(1+\frac1n\right)^n=e\tag{2} $$ where $n\in\mathbb{Z}$. Therefore, $(1)$ and the Squeeze Theorem says that $$ \lim_{\alpha\to\infty}\left(1+\frac1\alpha\right)^{\large\alpha}=e\tag{3} $$ where $\alpha\in\mathbb{R}$. Furthermore, $(3)$ implies $$ \begin{align} \lim_{\alpha\to\infty}\left(1-\frac1\alpha\right)^{\large-\alpha} &=\lim_{\alpha\to\infty}\left(1+\frac1{\alpha-1}\right)^{\large\alpha}\\ &=\lim_{\alpha\to\infty}\left(1+\frac1{\alpha-1}\right)^{\large\alpha-1}\left(1+\frac1{\alpha-1}\right)\\[9pt] &=e\cdot1\tag{4} \end{align} $$ Thus, $(3)$ and $(4)$ give $$ \lim_{|\alpha|\to\infty}\left(1+\frac1\alpha\right)^{\large\alpha}=e\tag{5} $$
Applying $\boldsymbol{(5)}$ to the Question $$ \begin{align} \lim_{n\to\infty}\left(1+\frac xn\right)^n &=\lim_{n\to\infty}\left(1+\frac xn\right)^{\large\frac nx\cdot x}\\[9pt] &=e^x\tag{6} \end{align} $$ for all $x$.